2.4 Applications of Uniform Convergence to Power Series

We want to apply the theorem on uniform convergence from the previous section to power series.

If \(\{ f_n\} _{n=0}^{\infty }\) is a sequence of complex-valued functions on a subset \(E\) of \(\mathbb {C}\), then the \(m\)th partial sum of \(\{ f_n\} \) is the function \(S_m:E\to \mathbb {C}\) given by \(S_m(z) = \sum _{n=0}^m f_n(z)\). We say that the series \(\sum _{n=0}^{\infty } f_n\) converges pointwise (resp. uniformly) on \(E\) to a function \(S:E\to \mathbb {C}\) if the sequence \(\{ S_m\} _{m=0}^{\infty }\) converges pointwise (resp. uniformly) to \(S\).

Theorem 2.25 Abel

Suppose \(z_0\in \mathbb {C}\) and \(\sum _{n\geq 0} a_n (z-z_0)^n\) is a power series with radius of convergence \(R\). And suppose that \(r\) is a real number such that \(0 {\lt} r {\lt} R\). Then \(\sum _{n\geq 0} a_n (z - z_0)^n\) converges uniformly on \(D(z_0, r)\).

Proof ▼

Assume that \(z_0 = 0\). (I’ll leave the rest of the argument for arbitrary \(z_0\) to the reader.) Pick \(\epsilon {\gt} 0\). We need to show that there exists an \(N\) such that, for all \(z\in D(0,r)\), \(|\sum _{n\geq N} a_n z^n | {\lt} \epsilon \).

Since \(r {\lt} R\), \(\sum _{n\geq 0} a_n r^n\) converges absolutely. So we can find an \(N\) such that \(\sum _{n\geq N} |a_n| r^n {\lt} \epsilon \). But then, if \(|z| {\lt} r\), \(|a_n z^n| \leq |a_n| r^n\). So

\begin{align*} |\sum _{n\geq N} a_n z^n| & \leq \sum _{n\geq N} |a_n z^n|\\ & \leq \sum _{n\geq N} |a_n| r^n {\lt}\epsilon . \end{align*}

Corollary 2.26

Suppose \(\sum _{n\geq 0} a_n (z-z_0)^n\) is a power series with radius of convergence \(R\). Then

\(\sum _{n\geq 0} a_n (z-z_0)^n\) converges on \(D(z_0,R)\) to an analytic function \(f\).
\(\sum _{n\geq 0} n a_n (z-z_0)^{n-1}\) converges on \(D(z_0,R)\) to \(f'\).
More generally, for any positive integer \(k\), \(\sum _{n\geq 0} n(n-1)\cdots (n-k+1) (z-z_0)^{n-k}\) converges on \(D(z_0,R)\) to \(f^{(k)}\).

Moreover, all the above convergences are uniform on \(D(z_0,r)\) for any positive real number \(r\) with \(r{\lt}R\).

Proof ▼

Apply Theorem 2.24.

Corollary 2.27

Suppose \(\sum _{n\geq 0} a_n (z - z_0)^n\) is a power series with radius of convergence \(R {\gt} 0\) converging to a function \(S\). Then \(\sum _{n\geq 0} a_n (z - z_0)^n\) is the Taylor series of \(f\).

Proof ▼

We already know that \(S\) is analytic. So it has a Taylor series.

Write \(S_n = \sum _{k=0}^n a_k (z - z_0)^k\) for the \(n\)th partial sum of the series. Then \(S_n^{(k)}\to S^{(k)}\) as \(n\to \infty \). So \(a_k = S_n^{(k)}(0)/k! \to S^{(k)}(0)/k!\).

Lemma 2.28

Suppose \(z_0\in \mathbb {C}\) and \(C\) is a simple closed curve in \(\mathbb {C}\) containing \(z_0\) in its interior. Let \(n\in \mathbb {Z}\). Then

\[ \frac{1}{2\pi i}\int _C (z - z_0)^n\, dz = \begin{cases} 1, & n = -1;\\ 0, & \text{else.} \end{cases} \]

Proof ▼

We can assume that \(z_0 = 0\) and that \(C\) is the unit circle centered at \(0\). Then we can parametrize \(C\) by \(\gamma :[0,2\pi ]\to \mathbb {C}\) given by \(t\mapsto e^{it}\). We get

\begin{align*} \frac{1}{2\pi i}\int _C z^n & = \frac{1}{2\pi i}\int _0^{2\pi } ie^{int}\cdot ie^{it}\, dt\\ & = \frac{1}{2\pi }\int _0^{2\pi } e^{i(n+1)t}\, dt\\ \end{align*}

If \(n = -1\), the integrand in the last integral is \(1\). So the integral is \(1\) and the Lemma is proved. If \(n \neq 1\), then

\[ \int _0^{2\pi } e^{i(n+1)t}\, dt = \frac{e^{i(n+1)t}}{n+1}\Bigm |_0^{2\pi } = 0. \]

Theorem 2.31

Suppose the Laurent series \(\sum _{n\in \mathbb {Z}} a_n (z - z_0)^n\) converges in an annular domain \(D\) centered at \(z_0\) to a function \(f\). Then \(f\in A(D)\) and \(\sum _{n\in \mathbb {Z}} a_n (z - z_0)^n\) is the Laurent series expansion of \(f\).

Sketch ▼

We can assume that \(z_0 = 0\) and that \(D = \{ z\in \mathbb {C}: r {\lt} |z| {\lt} R\} \) where \(0\leq r {\lt} R\leq \infty \). The same idea used to prove that power series converge uniformly shows that \(\sum _{n\geq 0} a_n (z - z_0)^n\) and \(\sum _{n {\lt} 0} a_n (z-z_0)^n\) both converge uniformly on any domain \(D'\) for the form \(D' = \{ z: r' {\lt} |z| {\lt} R'\} \) where \(r {\lt} r' {\lt} R' {\lt} R\). But then pick \(C\subset R'\) a simple closed curve with \(z_0\) in its interior. Now we can use Cauchy’s formula for the derivative of \(f\) along with the Lemma.